An electric field of 20 N C ⁻¹ is present along the x -axis in space. A charge of -2.0 10⁻⁴ C is transported from point A to point B . Determine the change in electrical potential energy, U_B - U_A , for the cases (a), (b) and (c) where the coordinates of A and B are given by: (a) A = (0, 0) ; B = (4 m , 2 m ) (b) A = (4 m , 2 m ) ; B = (6 m , 5 m ) (c) A = (0, 0) ; B = (6 m , 5 m )

Step-by-step solution

The change in electrical potential energy is given by the negative of the work done by the electric field. U = -W = -q E r Given E = 20 i N C ⁻¹ and q = -2.0 10⁻⁴ C . U = -(-2.0 10⁻⁴) (20 i ) ((x_B - x_A) i + (y_B - y_A) j ) U = 4.0 10⁻³ (x_B - x_A) J For case (a), A = (0, 0) and B = (4 m , 2 m ) : U_a = 4.0 10⁻³ (4 - 0) = 0.016 J For case (b), A = (4 m , 2 m ) and B = (6 m , 5 m ) : U_b = 4.0 10⁻³ (6 - 4) = 0.008 J For case (c), A = (0, 0) and B = (6 m , 5 m ) : U_c = 4.0 10⁻³ (6 - 0) = 0.024 J